Why do we write transition between two different electronic states like that: $$\bar p_{if}=\langle \psi_i|\bar p|\psi_f \rangle$$ And not like that: $$\bar p_{if}=\langle \psi_f|\bar p|\psi_i \rangle$$ where $i$ and $f$ mean initial and final states. As I understand, we want to see how much those states overlap each other, but why are we using operator on final state?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The two numbers are complex conjugate of each other - thus, whatever is the calculation, they are manipulated to give something real. E.g., in the at lowest order we simply use the Fermi Golden rule $$w_{i\rightarrow f} \propto |\langle \psi_i|\bar{p}|\psi_f\rangle|^2= \langle \psi_i|\bar{p}|\psi_f\rangle\langle \psi_i|\bar{p}|\psi_f\rangle^*= \langle \psi_i|\bar{p}|\psi_f\rangle\langle \psi_f|\bar{p}|\psi_i\rangle.$$
-
$\begingroup$ Okay, but according to fermi rule we should use operator on initial state $\endgroup$– coverCommented Oct 22, 2021 at 11:33
-
$\begingroup$ @cover the question is about the order of indices: it si convenient to have them written in the same order as in the bra and ket. Some field prefer to multiply operators in the other direction - it doesn't make difference, as long as it does not change the result. $\endgroup$– Roger V.Commented Oct 22, 2021 at 11:35